The number π (pi) has been studied since ancient times, and so has the concept of irrational numbers. An irrational number is any real number that cannot be expressed as a fraction a/b, where a is an integer and b is a non-zero integer.

In 1761, Lambert proved that π is irrational by first showing that this continued fraction expansion holds:

Then Lambert proved that if x is non-zero and rational then this expression must be irrational. Since tan(π/4) = 1, it follows that π/4 is irrational and therefore that π is irrational.[2] A simplification of Lambert's proof is given below. This result can also be proved using even more basic tools of calculus (integrals instead of series).[3][4]

This proof uses the characterization of π as the smallest positive number whose half is a zero of the cosine function and it actually proves that π2 is irrational.[5][6] As in many proofs of irrationality, the argument proceeds by reductio ad absurdum.

It follows by induction from this, together with the fact that A0(x) = sin(x) and that A1(x) = −x cos(x) + sin(x), that An(x) can be written as , where Pn and Qn are polynomial functions with integer coefficients and where the degree of Pn is smaller than or equal to ⌊n⁄2⌋. In particular,

Hermite also gave a closed expression for the function , namely

He did not justify this assertion, but it can be proved easily. First of all, this assertion is equivalent to

If π2/4 = p/q, with p and q in N, then, since the coefficients of Pn are integers and its degree is smaller than or equal to ⌊n⁄2⌋, q⌊n/2⌋Pn(π2/4) is some integer N. In other words,

But this number is clearly greater than 0; therefore, N ∈ N. On the other hand, the integral that appears here is not greater than 1 and

So, if n is large enough, N < 1. Thereby, a contradiction is reached.

Hermite did not present his proof as an end in itself but as an afterthought within his search for a proof of the transcendence of π. He discussed the differential-recurrent relations to motivate and to obtain the convenient integral representation. Once the integral is obtained, there are various ways to present a succinct and self-contained proof starting from the integral (as in Cartwright's or Niven's presentations), which Hermite could easily see (as he did in his proof of the transcendence of e[7]).

Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, An(x) is the "residue" (or "remainder") of Lambert's continued fraction for tan(x).[4]

This proof uses the characterization of π as the smallest positive zero of the sine function.[9]

Preparation: Suppose that π is rational, i.e. π = a /b for some integers a and b ≠ 0, which may be taken without loss of generality to be positive. Given any positive integer n, we define the polynomial function

and denote by

the alternating sum of f and its first n even derivatives.

Claim 1:F(0) + F(π) is an integer.

Proof: Expanding f as a sum of monomials, the coefficient of xk is a number of the form ck /n! where ck is an integer, which is 0 if k < n. Therefore, f (k)(0) is 0 when k < n and it is equal to (k! /n!) ck if n ≤ k ≤ 2n; in each case, f (k)(0) is an integer and therefore F(0) is an integer.

On the other hand, f(π – x) = f(x) and so (–1)kf (k)(π – x) = f (k)(x) for each non-negative integer k. In particular, (–1)kf (k)(π) = f (k)(0). Therefore, f (k)(π) is also an integer and so F(π) is an integer (in fact, it is easy to see that F(π) = F(0), but that is not relevant to the proof). Since F(0) and F(π) are integers, so is their sum.

which is smaller than 1 for large n, hence F(0) + F(π) < 1 for these n, by Claim 2. This is impossible for the positive integer F(0) + F(π).

The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula

which is obtained by 2n + 2integrations by parts. Claim 2 essentially establishes this formula, where the use of F hides the iterated integration by parts. The last integral vanishes because f (2n + 2) is the zero polynomial. Claim 1 shows that the remaining sum is an integer.

Niven's proof is closer to Cartwright's (and therefore Hermite's) proof than it appears at first sight.[4] In fact,

Proof: This can be proved by comparing the coefficients of the powers of x.

Claim 2: For each x ∈ R,

Proof: In fact, the sequence x2n/n! is bounded (since it converges to 0) and if C is an upper bound and if k > 1, then

Claim 3: If x ≠ 0 and if x2 is rational, then

Proof: Otherwise, there would be a number y ≠ 0 and integers a and b such that . In order to see why, take y = fk + 1(x), a = 0 and b = 1 if fk(x) = 0; otherwise, choose integers a and b such that fk + 1(x)/fk(x) = b/a and define y = fk(x)/a = fk + 1(x)/b. In each case, y cannot be 0, because otherwise it would follow from claim 1 that each fk + n(x) (n ∈ N) would be 0, which would contradict claim 2. Now, take a natural number c such that all three numbers bc/k, ck/x2 and c/x2 are integers and consider the sequence

Then

On the other hand, it follows from claim 1 that

which is a linear combination of and with integer coefficients. Therefore, each is an integer multiple of y. Besides, it follows from claim 2 that each (and therefore that gn ≥ ) if n is large enough and that the sequence of all 's converges to 0. But a sequence of numbers greater than or equal to cannot converge to 0.

Since f1/2(π⁄4) = cos(π⁄2) = 0, it follows from claim 3 that π2/16 is irrational and therefore π is irrational.

On the other hand, since

another consequence of claim 3 is that, if x ∈ Q\{0}, then tan x is irrational.

Laczkovich's proof is really about the hypergeometric function.[clarification needed] In fact, fk(x) = 0F1(k; −x2) and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation.[11] This allowed Laczkovich to find a new and simpler proof of the fact that the tangent function has the continued fraction expansion that Lambert had discovered.

Laczkovich's result can also be expressed in Bessel functions of the first kind . In fact, Γ(k)Jk − 1(2x) = xk − 1fk(x). So Laczkovich's result is equivalent to: If x ≠ 0 and if x2 is rational, then